Teacher2Teacher |
Q&A #9875 |
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Dear Pam, Here is an example of an alternate algorithm that helps students who are having difficulty with the traditional version. It helps students gain an understanding of what is happening during the division process, and assists those who do not have a good recall of all the multiplication facts, because they don't have to come up with the ONE correct digit for each step. Instead, they can use a succession of factors to get to the final quotient. Here is how it works... ____ 7 /478 7 X 10 the student can select any factor they wish to use - 70 to find a product to subtract from the original _____ amount (the dividend)-- usually a student 408 7 X 10 selects an easy-to-use factor like -70 1, 2, 5, or 10 ____ 338 7 X 10 -70 As students become more familiar with the ____ process they will select larger factors 268 7 X 10 -70 ____ 198 7 X 10 -70 ____ 128 7 X 10 -70 ___ 58 7 X 2 -14 ___ 44 7 X 2 -14 ___ 30 7 X 2 -14 ____ 16 7 X 2 When you have found all the "groups" of 7 that can be -14 subtracted out, you are finished. Now add all the ___ factors you multiplied by 7, and that will give you 2 the quotient. 10 + 10 + 10 + 10 + 10 + 10 + 2 + 2 + 2 + 2 = 68 See, this student was able to finish the problem without using any of the facts except 7 X 10 and 7 X 2. This could be of great assistance to someone who understood why we needed to divide, but didn't have the facts mastered yet. This is easier for struggling students because they don't have to have the perfect factor each time, just one that will work. They continue finding factors and dividing until they have a remainder that is smaller than the divisor (in this case, 7). When they have found all the factors, they just add them up (not the 7, but the other one each time). That is their quotient, and what is left is the remainder, which can be put in fraction form (in this case, 2 out of 7, or 2/7) , or they can place a decimal, and add zeroes to continue dividing. Besides that help, this process treats the dividend (in this case, 478) as a whole amount, instead of looking at 4 hundreds, then 48 tens, as the traditional algorithm does. For some students, that is a new way to look at the procedure. They have been used to looking at the amounts in terms of single digits... "How many 7's are in 4?" then "How many sevens are in 48?" No one has helped them see the big picture, that the 4 is really 4 hundreds, and the 48, 48 tens... I try to move my students to try larger "chunks" once we have that first step mastered... ___ 7 /478 7 X 30 -210 ____ 268 7 X 30 -210 ____ 58 7 X 5 -35 ____ 23 7 X 3 -21 ____ 2 30 + 30 + 5 + 3 = 68 Then we have a challenge to try to find the one "best chunk" for each place value. In the next example, there is one "chunk" for the tens, and one for the ones. (There is no hundreds chunk, because 7 X 100 would be more than 478). ___ 7 /478 7 X 60 -420 ____ 58 7 X 8 -56 ____ 2 60 + 8 = 68 You will probably notice that that is VERY similar to the traditional algorithm. It is just a small nudge away for most students, at that point. :-) -Gail, for the T2T service |
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